Conflict Between Time Dilation and Red/Blue Shift?

[Sagittarius A-Star]

No. The relations between the angles are of course different. You find the formulae for both Doppler effect and aberration, using the standard Lorentz boost for the plane em. wave field, in Sect. 3.2.1 here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The aberration formula I wrote is identical to that in Einstein 1905, §7, with $\beta$ in the source frame and the case that the angle in the source frame ist 90 degree. He also wrote $\cos {\alpha} = -v/c$.

It is the same as your formula (3.2.62) for $\alpha' = 90°$, except the sign of $\beta$.

 

[vanhees71]

The different sign comes simply from the different meaning of $\vec{v}$. In description by Einstein, the observer is moving with $\vec{v}_E$ in the rest-frame of the light source, while in my description $\vec{v}$ is the velocity of the light source in the rest frame of the observer, i.e., $\vec{v}=-\vec{v}_E$. Thus mine and his equations are equivalent.

 

[H_A_Landman]

One simple way to understand both of these cases qualitatively is to remember the Einstein equivalence principle that acceleration and gravity are locally equivalent, and that there is accelerational time dilation. First derived by Einstein in 1907 (using only SR), it says e.g. that a clock near the front of an accelerating rocket must run slightly faster than a clock near its tail.

Summary:: Does Red/Blue Shift Indicate Relative Passage of Time?

Experiment 1: Astronaut travels away from Earth at near the speed of light, then travels toward the Earth at near the speed of light.

What you are ignoring are the accelerations at the launch time and the turnaround time.

At launch time, while accelerating (let's say at 1 G), the astronaut imputes a time dilation field to the whole universe, with clocks running faster in the "up" (ahead of her) direction and slower in the "down" (behind her), and the size of the effect dependent on distance. Because the Earth is close, the effect is pretty small, and I'll neglect it here.

At turnaround time, the astronaut must accelerate toward Earth to slow down and stop, and an equal amount to speed up in the return direction. During this entire acceleration, Earth is "above" the astronaut, by roughly the full distance of travel, and so she perceives clocks on Earth to be running much faster than her own clocks. This is a big chunk of the final clock difference when she returns. Changing the G-force doesn't affect this much; if you doubled the acceleration, you'd have twice the effect for half the time, and get about the same total time shift.

Experiment 2: Astronaut circles the Earth at near the speed of light.

Again in this case, the astronaut is accelerating, and the Earth is in the "up" direction, so its clocks are running faster in her (non-inertial) frame.

 

[PeterDonis]

there is accelerational time dilation

This is a somewhat misleading way to put it. The time dilation is caused by a difference in height in a "gravitational field", not by acceleration. Acceleration can "create" a "gravitational field", but it is the difference in height in the field, not the acceleration itself, that causes the time dilation. (Note that in the standard "twin paradox", the stay-at-home twin, whose clock is supposed to be running very, very fast during the turnaround, is in free fall.)

Also, as my use of the quotes indicates, the "gravitational field" that is "created" by acceleration turns out to be a frame-dependent thing and not a good basis on which to construct a general theory. While it works qualitatively in these particular cases, it doesn't generalize well.

 

[H_A_Landman]

The time dilation is caused by a difference in height in a "gravitational field", not by acceleration. Acceleration can "create" a "gravitational field", but it is the difference in height in the field, not the acceleration itself, that causes the time dilation.

The difference in time rates is (to first order) proportional to both the height $h$ and the acceleration $g$, and thus to their product $gh$. Yes? The weak field approximation is $$T_d = 1 + gh/c^2$$ in which they both matter. So I don't understand your quibble here; it seems incorrect, or perhaps poorly stated. Are you saying that if you hold the acceleration constant then it's only a function of height? That's true but incomplete. If you hold the height constant, then it's only a function of the acceleration.

The goal here was to provide a simple qualitative Level I answer to show the questioner "where they went wrong". Ignoring the effects of the terminal accelerations is one way to go very wrong. (There are other ways.) Whether I succeeded or not should primarily be judged by how useful (or useless) my answer was to her.

Einstein's 1907 paper derives accelerational time dilation first (in section 18), and then using the EEP concludes that there must also be gravitational time dilation. The initial derivation doesn't depend on any theory of gravity at all; it only uses SR. Today we would tend to start with GR and work downwards, but that's not required. It doesn't depend on GR, only SR. The existence and (first order) magnitude of gravitational time dilation doesn't depend on GR either; it's already forced by SR + EEP.

 

[Dale]

The difference in time rates is (to first order) proportional to both the height h and the acceleration g, and thus to their product gh. Yes?

Yes, that is why he said “difference in height in the field”. I prefer to say “potential” or “gravitational potential” since that specifies both the dependence on the height and the field.

Einstein's 1907 paper derives accelerational time dilation first (in section 18), and then using the EEP concludes that there must also be gravitational time dilation.

Except that even in an accelerating frame the time dilation is not attributable to acceleration. The clock hypothesis, which has been experimentally confirmed to 10^18 g, states that the accelerational time dilation is 0.

Time dilation, both in an accelerating frame or a gravitational field is attributed to differences in gravitational potential, not to acceleration. Calling it accelerational time dilation is bad semantics and worse pedagogy and I strongly recommend against it.

 

[PeterDonis]

The difference in time rates is (to first order) proportional to both the height $h$ and the acceleration $g$, and thus to their product $gh$. Yes?

Yes, but the "acceleration" here is the acceleration of the frame, not of any particular object. An object in free fall at height $h$ still has the same time dilation.

 

[H_A_Landman]

... even in an accelerating frame the time dilation is not attributable to acceleration. The clock hypothesis, which has been experimentally confirmed to 10^18 g, states that the accelerational time dilation is 0.

I never claimed that it wasn't, and I wasn't talking about that (non-existent) effect at all. (In fact Einstein 1907 assumes that it is zero; it's part of the clock synchronization in the derivation.) Still, all that says is that if $h$ is zero, $gh$ is also zero, with which I completely agree. But it's equally true that if $g$ is zero, $gh$ is zero. There's no positional time dilation if there's no acceleration. It is not a function only of position, which many of your statements appear to claim. We have a function of 2 variables, which (except for units) is symmetric in the 2 variables; so when you say that the function is entirely due to 1 of the variables and is "not attributable" to the other variable, that sounds very strange to me. It denies the symmetry.

I think we agree on what the effect is (and isn't), and you're mainly objecting to my use of "accelerational" to describe it. What term would you suggest instead? "Potential-based"? I might be OK with that, even though in the original question we're talking about SR, so there is no gravity, and it's only a pseudo-potential generated by the acceleration of the astronaut's frame.

I completely agree with your "acceleration of the frame ... free fall" comment.

 

[PeterDonis]

we're talking about SR, so there is no gravity, and it's only a pseudo-potential generated by the acceleration of the astronaut's frame

The same is true in GR; what you are calling "gravity" is an artifact of using a non-inertial frame.

 

[Dale]

It is not a function only of position, which many of your statements appear to claim.

I think that you are confusing me with @PeterDonis

you're mainly objecting to my use of "accelerational" to describe it.

Yes. That is my objection. I also have never seen the term “accelerational time dilation” used in the scientific literature.

What term would you suggest instead? "Potential-based"?

I personally would prefer “gravitational time dilation” since that is the term I have seen used in the scientific literature. But if you are going to use non-standard terminology then, yes, “potential-based” or something similar would be far more accurate.

I cannot tell you how many people on these forums have the mistaken belief that gravitational time dilation is due to $g$ and not $gh$. Calling it “accelerational” would certainly exacerbate that.

even though in the original question we're talking about SR, so there is no gravity, and it's only a pseudo-potential generated by the acceleration of the astronaut's frame

The equivalence principle and Einstein’s usage certainly justifies calling such a field gravity. Furthermore, even without invoking relativity at all the pseudo-potential is a full fledged potential. In those coordinates, the work done about a closed loop is zero, the gradient gives a force, and in those coordinates there is a corresponding term in the Lagrangian with a related symmetry.

However, if you do object to the use of “potential” because of the fact that this potential only shows up in certain coordinate systems, please be aware that the same is true of gravitational time dilation in curved spacetime.

It denies the symmetry.

As does calling it “accelerational”

 

[PeterDonis]

It is not a function only of position, which many of your statements appear to claim.

Yes, as far as your formula is concerned, it is, because, as I have already pointed out, the $g$ in your formula is not a coordinate, as $h$ is. The $g$ is a property of the frame you have chosen, and it is a constant everywhere in that frame, so nothing can be a function of it.

 

[PeterDonis]

if you do object to the use of “potential” because of the fact that this potential only shows up in certain coordinate systems, please be aware that the same is true of gravitational time dilation in curved spacetime.

Actually, in a stationary spacetime, "gravitational potential" can be defined in a coordinate independent way using the timelike Killing vector field. For the Rindler congruence in flat spacetime, which is what is used in the "accelerating rocket" scenario, the timelike KVF is the "boost" KVF, for which the worldlines of the Rindler congruence are integral curves; so that KVF can work just fine to define a gravitational potential independent of coordinates. In Schwarzschild spacetime, a similar construction can be done using the static worldlines (worldlines of observers "hovering" at a constant altitude). (In a spacetime that is not stationary, no useful notion of "gravitational potential" can be defined at all.)

However, the $g$ in the $gh$ that @H_A_Landman is talking about is not the gravitational potential in any sense.

 

[Sagittarius A-Star]

The $g$ is a property of the frame you have chosen, and it is a constant everywhere in that frame, so nothing can be a function of it.

I don't understand, how this can be consistent to equation (5) in the following paper:
https://arxiv.org/abs/gr-qc/0409033

I agree, that nothing can be a function of the constant $g$ in the formula of @H_A_Landman.

 

[Dale]

Actually, in a stationary spacetime, "gravitational potential" can be defined in a coordinate independent way using the timelike Killing vector field.

Yes. But what is frame invariant in terms of time dilation is the total redshift. The decomposition of that into gravitational and kinematic time dilation is always coordinate dependent, even for a stationary spacetime.

 

[PeterDonis]

what is frame invariant in terms of time dilation is the total redshift. The decomposition of that into gravitational and kinematic time dilation is always coordinate dependent, even for a stationary spacetime.

No, this is not correct. In a stationary spacetime, there is an invariant notion of "being at rest"--following an integral curve of the timelike KVF. Each worldline of the timelike KVF marks a "position" in an invariant sense. That means there is an invariant way to distinguish "gravitational" and "kinetic" time dilation; the "gravitational" part is due to a difference in position, and the kinetic part is due to velocity with respect to observers at rest.

It is true that this construction only works in stationary spacetimes, which causes a lot of confusion when people try to generalize it to non-stationary spacetimes like FRW spacetime and find that it doesn't work.

 

[Dale]

That means there is an invariant way to distinguish "gravitational" and "kinetic" time dilation; the "gravitational" part is due to a difference in position, and the kinetic part is due to velocity with respect to observers at rest.

Hmm, I haven’t thought about it that way. I see what you are saying, but I have never actually seen kinematic time dilation formulated in that way, as an invariant.

I think that I disagree about calling the invariant gravitational quantity “gravitational time dilation”. In my usage, time dilation is always the ratio between coordinate time and proper time which is inherently coordinate dependent. So the invariant quantity you are describing I think I would call gravitational redshift.

I agree that a frame invariant gravitational redshift and a frame invariant total redshift together imply a frame invariant kinematic redshift, which is something I haven’t considered before.

 

[PeterDonis]

I don't understand, how this can be consistent to equation (5) in the following paper

The "acceleration" in that paper is the proper acceleration of an observer at a given height. But that is not the same as the $g$ that is a property of the frame. In the paper, Alice has a proper acceleration $g^-$ and Bob has a proper acceleration $g^+$. But that does not mean that Alice's time dilation is $1 + g^- h$ and Bob's is $1 + g^+ h$. As the paper shows (Equation 26), time dilation of any object at height $h$ relative to Alice, including Bob, is $1 + g^- h$--so $g^-$ is the $g$ in Alice's (non-inertial) rest frame. A similar calculation would show that $g^+$ is the $g$ in Bob's (non-inertial) rest frame. But once you've picked a frame, you use the same $g$ for all time dilations in that frame; you don't switch from one $g$ to another depending on which object's time dilation you are calculating.